Ta có \(\widehat{BAH}=\widehat{ACB}\left(cùng.phụ.\widehat{B}\right)\);\(\widehat{CAH}=\widehat{CBA}\left(cùng.phụ.\widehat{ACB}\right)\)
\(\Rightarrow\widehat{IAH}=\widehat{IAB}=\widehat{ACI}=\widehat{BCI}\left(\dfrac{1}{2}\widehat{BAH}=\dfrac{1}{2}\widehat{ACB}\right)\)
Ta có \(\widehat{CIA}=180-\widehat{CAI}-\widehat{ACI}=180-\widehat{CAH}-\widehat{IAH}-\widehat{IAH}=180-\left(\widehat{ABC}+\widehat{HAB}\right)=180-\left(\widehat{ABC}+\widehat{ACB}\right)=180-90=90\)
\(\widehat{AIC}=180^0-\widehat{ICA}-\widehat{IAC}\)
\(=180^0-\dfrac{\widehat{ACH}+\widehat{ACH}}{2}\)
\(=90^0\)
